A337832 Porous numbers: Numbers k which are not multiples of 10 such that every m with sum of digits = k and k a divisor of both m and rev(m) has a zero in its digits.

11, 37, 74, 101, 121

Offset

1,1

Comments

It is straightforward to prove that a number is not porous by finding an m that does not contain a zero, but satisfies the three requirements. For example, 12 is not a term because 48 is a number with sum of digits equal to 12 and both 48 and 84 are multiples of 12. Many numbers of A333666 can be taken directly to prove that a given value is not a term.

So far only for the five numbers of this sequence no m could be found by computer programs to demonstrate that they are not porous. For a good reason, because for these 5 numbers it can be proven that no such m exists (the proofs are provided in the link below).

Up to k = 1000 no further porous numbers exist (see the linked file with the corresponding m where k | m and k | rev(m)). The steadily increasing number of possibilities to construct an m for a given k suggests that the 5 terms might be the only porous numbers, but a mathematical proof for this conjecture seems a big challenge.

Links

Table of n, a(n) for n=1..5.

Ruediger Jehn, Numbers m that disprove k is porous

Ruediger Jehn, Proof for the first 5 terms

Rüdiger Jehn, Porous Numbers, arXiv:2104.02482 [math.GM], 2021.

Example

If we search for an m to demonstrate that k = 11 is not porous, we loop through all m where sum of digits = 11 and 11 a divisor of both m and rev(m). We find 209, 308, 407, ... 902, 2090, 3080, ... All m contain at least one zero. If it can be proven that this holds for all m, then 11 is a porous number.
For k = 11 this proof actually is quite easy:
Let "m_s ... m_3 m_2 m_1 m_0" be a number m with digits m_i and the sum of the digits is 11. We define:
A = m_0 + m_2 + m_4 + ... and B = m_1 + m_3 + m_5 + ...
A divisibility rule for 11 requires that the alternating sum of the digits must be a multiple of 11. Hence:
   A - B = j * 11
Since the sum of the digits is 11, we have
   A + B = 11
Adding the two equations yields
   2 * A = (j + 1) * 11
Therefore A must be 0 or 11. If A is 11, then B is 0. This means either A or B must be zero and m must contain a zero. Hence 11 is a porous number.

Crossrefs

Cf. A333666.

Sequence in context: A358079 A160023 A263201 * A188135 A188382 A090950

Adjacent sequences: A337829 A337830 A337831 * A337833 A337834 A337835

Keyword

nonn,base,more

Author

Ruediger Jehn, Sep 24 2020

Status

approved